bài 1 : Tìm x thuộc Z, biết
1) \(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\) 2) \(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)
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a)\(\frac{-5}{6}\).\(\frac{120}{25}\)<x<\(\frac{-7}{15}\).\(\frac{9}{14}\)
-4 <x<\(\frac{-3}{10}\)
\(\frac{-40}{10}\)< x <\(\frac{-3}{10}\)=>x E {-39:-38:-37:.....:-4}
b)\(\left(\frac{-5}{3}\right)^3\)<x<\(\frac{-24}{35}.\frac{-5}{6}\)
\(\frac{-875}{189}< x< \frac{108}{189}\)
=> x E {\(\frac{-874}{189},\frac{-873}{189},......,\frac{107}{189}\)}
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
Bài 1: Tìm x biết:
1) x +\(\frac{7}{12}\)= \(\frac{17}{18}\)- \(\frac{1}{9}\) 2) \(\frac{29}{30}\)- (\(\frac{13}{23}\)+ x) = \(\frac{7}{69}\)
x +\(\frac{7}{12}\)= \(\frac{15}{18}\) \(\frac{13}{23}\)+ x = \(\frac{29}{30}\)- \(\frac{7}{69}\)
x = \(\frac{15}{18}\)- \(\frac{7}{12}\) \(\frac{13}{23}\)+ x = \(\frac{199}{230}\)
x = \(\frac{1}{4}\) x = \(\frac{3}{10}\)
1
Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
1) \(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\Leftrightarrow\frac{-5}{6}.\frac{24}{5}< x< \frac{-63}{210}\)
\(\Leftrightarrow-40< x< \frac{-63}{210}\)
\(\Leftrightarrow\frac{-400}{10}< \frac{10x}{10}< \frac{-3}{10}\)
\(\Leftrightarrow-400< 10x< -3\)
\(\Leftrightarrow x\in\left\{-39;-38;...;-2;-1\right\}\)
2) \(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)
\(\Leftrightarrow\frac{-125}{25}< x< \frac{4}{7}\)
\(\Leftrightarrow\frac{-35}{7}< \frac{-7x}{7}< \frac{4}{7}\)
\(\Leftrightarrow-35< -7x< 4\)
\(\Leftrightarrow x\in\left\{4;3;2;1;0\right\}\)